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skad [1K]
3 years ago
14

I need some help please

Mathematics
1 answer:
Nataliya [291]3 years ago
4 0

Answer:

25π

Step-by-step explanation:

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need help.. The midpoint of GH is M(-6,-3). One endpoint is H (-4,4) Find the coordinates of endpoint G.
iren2701 [21]

Answer:

The coordinates of G are: (-8, -10)

Step-by-step explanation:

Mid point formula:

x = \dfrac{x_1+x_2}{2}\\y = \dfrac{y_1+y_2}{2}

Where,

(x,y) are the coordinates of the mid point of the points (x_1,y_1)  and (x_2,y_2).

As per question statement, we are given that:

x_2=-4\\y_2=4

and

x=-6\\y=-3

To find:

x_1 = ?\\y_1 = ?

Solution:

Let us use the mid point formula:

-6 = \dfrac{x_1-4}{2}\\\Rightarrow -12=x_1-4\\\Rightarrow x_1=-12+4\\\Rightarrow \bold{x_1=-8}

-3 = \dfrac{y_1+4}{2}\\\Rightarrow -6=y_1+4\\\Rightarrow y_1=-6-4\\\Rightarrow \bold{y_1=-10}

The coordinates of G are: (-8, -10)

5 0
3 years ago
The price of a train ticket from Orlando to Atlanta is normally $118.00. However, the train company is offering a special 75% di
pogonyaev

Answer:

$88.50

Step-by-step explanation:

118*0.75 = 88.50

8 0
3 years ago
0.08 Divided by 3.84
Cloud [144]
Around 0.02 or more precise, 0.02083
8 0
3 years ago
Read 2 more answers
The heights of women in the USA are normally distributed with a mean of 64 inches and a standard deviation of 3 inches.
Rainbow [258]

Answer:

(a) 0.2061

(b) 0.2514

(c) 0

Step-by-step explanation:

Let <em>X</em> denote the heights of women in the USA.

It is provided that <em>X</em> follows a normal distribution with a mean of 64 inches and a standard deviation of 3 inches.

(a)

Compute the probability that the sample mean is greater than 63 inches as follows:

P(\bar X>63)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{63-64}{3/\sqrt{6}})\\\\=P(Z>-0.82)\\\\=P(Z

Thus, the probability that the sample mean is greater than 63 inches is 0.2061.

(b)

Compute the probability that a randomly selected woman is taller than 66 inches as follows:

P(X>66)=P(\frac{X-\mu}{\sigma}>\frac{66-64}{3})\\\\=P(Z>0.67)\\\\=1-P(Z

Thus, the probability that a randomly selected woman is taller than 66 inches is 0.2514.

(c)

Compute the probability that the mean height of a random sample of 100 women is greater than 66 inches as follows:

P(\bar X>66)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{66-64}{3/\sqrt{100}})\\\\=P(Z>6.67)\\\\\ =0

Thus, the probability that the mean height of a random sample of 100 women is greater than 66 inches is 0.

8 0
3 years ago
Mr. Gonzalez has money in three different savings accounts. One pays 2.3% interest. The second offers 2.9% interest, and the thi
BlackZzzverrR [31]

So, I believe your correct answer is C. 2.3

5 0
3 years ago
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